# Inner product vs scalar rpoduct

We define an inner product similar to a scalar product:

\begin{align*} \text{scalar product:}& & \vec{a}\cdot\vec{b}&= a_1b_1 + a_2 b_2 + a_3b_3 + \dots\\ \text{inner product:}& & \langle a | b \rangle &= \overline{a_1}b_1 + \overline{a_2}b_2 + \overline{a_3}b_3+\dots \end{align*}

So the inner product is designed to work in $\mathbb{C}$ similar to how scalar product works in $\mathbb{R}$. So the question arizes about a geometrical interpretation of an inner product. For scalar product it is $|\vec{a}||\vec{b}|\cos{\varphi}$ which we interpret as:

˝Projection of a 1st vector's norm to a 2nd vector multiplied by a 2nd vector's norm.˝

Question: What about for an inner product? Does the above quotation hold?

Some background on my question: I am asking this because in many quantum mechanical books they are talking of a state vector $\left|\Psi(t)\right\rangle$ which is an abstraction that defines the state of a system and is only a function of time $t$. Then we have to define a base whose base vectors and the space (we talk about position, momentum space...) we get is dependant of the base vectors we choose. So if i am talking about position space in 1-D some physicists claim that this inner product:

$$\left\langle x \right.\left|\Psi(t)\right\rangle$$

is a projection of a state vector $\left| \Psi(t)\right\rangle$ onto a normalised position base vector $\left|x\right\rangle$ and we can interpret this as $\Psi(x,t)$ which is now a function of position!

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It's better not to use the term "scalar product" at all. The first thing is a real inner product and the second thing is a complex inner product. If you want to refer to the first thing where the $a_i$ and $b_i$ are complex, call it a symmetric bilinear form. –  Qiaochu Yuan May 26 '13 at 9:40
Thanks on the note! –  71GA May 26 '13 at 10:09
@QiaochuYuan: unfortunately it is not that easy to kill off a French tradition. –  Willie Wong May 26 '13 at 10:58
Well thats how they taught us in schools. –  71GA May 26 '13 at 11:07